Question

If $\vec{A}=\hat{i}+2\hat{j}-\hat{k},\vec{B}=-\hat{i}+\hat{j}-2\hat{k}$, then angle between $\vec{A}$ and $\vec{B}$ is

• A. $\frac{\pi }{2}$
• B. 0
• C. $\pi$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (D)

$\vec{A}=\hat{i}+2\hat{j}-\hat{k};\vec{B}=-\hat{i}+\hat{j}-2\hat{k}$
$A=\sqrt{1^2+2^2+(-1{)}^2}=\sqrt{6};B=\sqrt{(-1{)}^2+1^2+(-2{)}^2}=\sqrt{6}$
$\vec{A}\cdot \vec{B}=(\hat{i}+2\hat{j}-\hat{k})\cdot (-\hat{i}+\hat{j}-2\hat{k})=-1+2+2=3$
Let $\theta$ be the angle between $\vec{A}$ and $\vec{B}$. Then
$\vec{A}\cdot \vec{B}=AB\cos \theta$
or $3=\sqrt{6}\sqrt{6}\cos \theta$
or $\cos \theta =\frac{3}{6}=\frac{1}{2}$
or $\theta =\frac{\pi }{3}$